Eduqas A-Level Physics A420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Conduction of Electricity
Resistance
D.C. Circuits
Capacitance
Solids Under Stress
Electrostatic and Gravitational Fields of Force
Using Radiation to Investigate Stars
Orbits and the Wider Universe
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Nuclear Decay
Particles and Nuclear Structure
Nuclear Energy
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Energy Concepts
Power, dissipative forces and efficiency
- 3 things to know
- Component 1
Overview
Equation P = Fv, defining every symbol and its unit, dissipative force and equation for efficiency as a percentage.
What you need to know
Answer each one before you reveal it.
State the equation P = Fv, defining every symbol and its unit, and explain how it follows from P = W/t.
Explain why a car has a maximum speed on a level road even though the engine power is constant.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Define power.
Model answer
Power is the rate at which energy is transferred, or equivalently the rate at which work is done.
- 2
State the equation linking power, work and time, defining every symbol and its unit.
Model answer
The equation is P = W/t, where P is the power in W, W is the work done or energy transferred in J and t is the time taken in s.
- 3
Define the watt.
Model answer
One watt is a rate of energy transfer of one joule per second.
- 4
State the equation P = Fv, defining every symbol and its unit, and explain how it follows from P = W/t.
Model answer
In P = Fv, P is the power in W, F is the force in N and v is the steady speed in m s⁻¹. Since W = Fx, dividing by t gives P = F(x/t), and x/t is the speed.
- 5
A crane raises a 500 kg load through 12.0 m in 20.0 s. Calculate the useful power output. Take g as 9.81 m s⁻².
Model answer
The work done is 500 × 9.81 × 12.0 = 5.89 × 10⁴ J, so the power is 58860 ÷ 20.0 = 2.94 × 10³ W. Multiplying 2943 W by 20.0 s returns 5.89 × 10⁴ J.
- 6
A car engine delivers 30.0 kW while the car travels at a steady 20.0 m s⁻¹. Calculate the total resistive force.
Model answer
At constant speed the driving force equals the resistance, so F = 30000 ÷ 20.0 = 1.50 × 10³ N. Multiplying 1500 N by 20.0 m s⁻¹ returns 30.0 kW.
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